It's been decades since I looked at the Centripetal Force equation. Something like F=-wr^2 So to create artificial gravity during space travel, the space craft must spin at w radians/sec. But the gravity would only be felt at the walls of the spinning spacecraft. As one approached the axis of rotation, the effects would be diminished. In fact, if you were to be standing upright and holding a ball the ball would not fall to your feet as you would expect it because if you release the ball. It would experiene zero gravity until the walls of the spacecraft reaches the free-floating ball. In a spinning spacecraft, you would still be able to jump from floor to ceiling (landing on your head).
it depends http://www.permanent.com/s-centri.htm it all depends on how much gravity you want and how big your wheel is.
good question: Just to nitpick (adding nothing to the physics). There is never 'artificial' gravity. There is only creating conditions to use forces to simulate other environments. The actual gravitational forces are as real as in in any other situation.
explain gravity? Annyway earths gravity is diminished in orbit but far from gone, the astronauts in the space station aren't floating because they gravity is gone but because their in free fall. I'm to lazy to do the math but I'm sure there is more then 90% of earths gravity at the height of the ISS
Please Register or Log in to view the hidden image! http://www.nas.nasa.gov/About/Education/SpaceSettlement/75SummerStudy/Chapt4.html
I don't think you're correct here. If you're holding a ball and drop it, it will want to continue moving in a straight line which will cause it to move toward the floor although in a slightly curving trajectory. It would not float as in zero g. Saying it would experience no "gravity" (centripital force) until it reaches the wall is like saying a skydiver experiences no gravity until he hits the ground!
tostig: The force on a mass m travelling in a circle is: \(F=m\omega^2 r = mv^2/r\) where \(\omega\) is the angular speed and v is the linear speed. r is radius of the circle. madanthonywayne's comments above are correct. You are wrong to say that the ball will not fall at your feet when you drop it. The reason is that you ignore the initial velocity of the ball. As you hold it, in your rotating station, it has the same tangential (sideways) velocity as you do. When you release it, it keeps travelling at the same speed tangentially, but you and the station move "upwards" away from it. From your point of view, it appears to fall directly to the floor. Not unless the rate of spin was low enough, for the same reason as above.
This is a slight oversimplification, as the ball will fall slightly off to one side because of the rotation effects. This effect is less noticeable in larger rotating habitats. Here is an image that shows how the trajectory of a falling object differs in different rotating situations; Please Register or Log in to view the hidden image! As you can see, a falling object is displaced slightly to one side; similarly a hopping person is displaced as well. The larger the rotating vessel, the lower the r.p.m. required to acheive any particular gravity; and the smaller these effects wil be.
Eburacum45: A very good post. When we make a ship of 10 kilometers in diameter, only the most astute observer will recognize your logic!
:bugeye: is that sentence correct? if you find 10 km absurd rest asure it doesn't have to be that long and you don't even have to have a ring structure a 10 km tether and a ruble pile on the other side could do the trick